Aptitude - Calendar - Discussion

Hey, this type of questions can be easily done by following formula
---------------------------------------------------------------------------------
f= K+[(13m-1)/5]+ d+[d/4]+[c/4]-2c

Here k= date of which you have to calculate(15 in above que.)
m= month (for march, m=1 and so on..for jan m=11)
d= last two digit of given year(10 in above que.)
c= first two digit of given year(20 in above que.)
---------------------------------------------------------------------------------
note: take only integer value after division for ex. 13/5=2
---------------------------------------------------------------------------------

For given question

15th August, 2010?

f=15+[(13*6-1)/5]+10+[10/4]+[20/4]-2*20
=15+15+ 10 + 2 + 5 -40
=7

Now 7%7==0

So its SUNDAY.

How for march the value of month is 6? I don't understand the month value.

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days.

How com tat 4 odd days ?

@vishnu : those 4 days are the result of 11days(divide these 11days into weeks and days, so u get 7+4,here 7days makes a week and do count those 4 extra days.) so finally we get 4ODD DAYS..

Dear friends,

Pls. explain which day will come on 25th dec 1992 with step by step procedure

Friends can you explain 7%7=0 taken as sunday how?

Why you are taken one time 1600, 400 or 1600, 300 in sum it's not possible to me to understand please help me?

In 2nd and 3rd step, from where does that 1600 and 400 years came?

How to find odd years in 100 years, 80 years or so.... Is their any formula to find it rather to go year by year calculation.

How can I know 9 years = (7 ordinary years +2 leap years).